Problem: $\dfrac{ -2r + 4s }{ -6 } = \dfrac{ -10r - t }{ 6 }$ Solve for $r$.
Solution: Notice that the left- and right- denominators are opposite $\dfrac{ -2r + 4s }{ -{6} } = \dfrac{ -10r - t }{ {6} }$ So we can multiply both sides by $-6$ $-{6} \cdot \dfrac{ -2r + 4s }{ -{6} } = -{6} \cdot \dfrac{ -10r - t }{ {6} }$ $-2r + 4s = - \cdot \left( -10r - t \right) $ Distribute the negative sign on the right side. $-2r + 4s = 10r + t$ $-{2}r + {4}s = {10}r + {1}t$ Combine $r$ terms on the left. $-{2r} + 4s = {10r} + t$ $-{12r} + 4s = 1t$ Move the $s$ term to the right. $-12r + {4s} = 1t$ $-12r = t - {4s}$ Isolate $r$ by dividing both sides by its coefficient. $-{12}r = t - 4s$ $r = \dfrac{ t - 4s }{ -{12} }$ Swap signs so the denominator isn't negative. $r = \dfrac{ -{1}t + {4}s }{ {12} }$